Given Information:
Mass of sock = 0.23 kg
Stretched length of sock = x = 2.54 cm = 0.0254 m
Required Information:
Spring constant = k = ?
Answer:
Spring constant = k = 88.82 N/m
Explanation:
We know from the Hook's law that
F = kx
Where k is spring constant, F is the applied force and x is length of sock being stretched.
k = F/x
Where F is given by
F = mg
F = 0.23*9.81
F = 2.256 N
So the spring constant is
k = 2.256/0.0254
k = 88.82 N/m
Therefore, the spring constant of the sock is 88.82 N/m
Answer:
They preserve many good_from historyof nepal_for us(from history of nepal)
A : lighter than lead shot, reducing its velocity and distance
Answer:
<h2>B. 20°</h2>
Explanation:
Range in projectile is defined as the distance covered in the horizontal direction. It is expressed as R = U²sin2Ф/g
U is the initial velocity of the body (in m/s)
Ф is the angle of projection
g is the acceleration due to gravity.
Given U = 14m/s, g = 9.8m/s and range R = 15 m
we will substitute this value into the formula to get the projection angle Ф as shown;
15 = 15²sin2Ф/9.8
15*9.8 = 15²sin2Ф
147 = 225sin2Ф
sin2Ф = 147/225
sin2Ф = 0.6533
2Ф = sin⁻¹0.6533
2Ф = 40.79°
Ф = 40.79°/2
Ф = 20.39° ≈ 20°
Hence, the range is greatest at angle 20°
Answer:
Explanation:
As it is given that flow rate in the pipe is 20 cm^3/s
so we have
at the upper end the area is given as
Also at the other end
now the speed at two ends is given as
now by Bernoulli's theorem we have
now we have
Now we have
